\documentclass[a4paper]{article}
\usepackage[margin=1in]{geometry}
\usepackage{ctex}
\usepackage{tikz}
\usepackage{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{fontspec}
\usepackage{amsthm}
\usepackage{xltxtra}
\usepackage{mflogo,texnames}
\usepackage{graphicx}
\usepackage{titlesec}

\setmainfont{Times New Roman}
\setCJKmainfont[BoldFont=SimHei,ItalicFont = SimSun]{SimSun}

\newtheorem{definition}{Definition}[section]%定义
\newtheorem{theorem}{Theorem}[section]%定理
\newtheorem{axiom}{Axiom}[section]%公理
\newtheorem{lemma}{Lemma}[section]%引理
\newtheorem{proposition}{Proposition}[section]%命题
\newtheorem{corollary}{Corollary}[section]%推论
\newtheorem{remark}{Remark}[section]%注

\title{\heiti\zihao{2} 习题4.4}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{求下列函数的二阶导数:}
\subsection{$y=x \sqrt{1-x^{2}}$}
\textbf{解}\quad
$$y^{\prime}=\sqrt{1-x^{2}}+\left(\frac{-x^{2}}{\sqrt{1-x^{2}}}\right)=\frac{-2 x^{2}+1}{\sqrt{1-x^{2}}},y^{\prime \prime}=\frac{2 x^{3}-3 x}{\left(1-x^{2}\right)^{\frac{3}{2}}}$$

\subsection{$y=\mathrm{e}^{-x^{2}}$}
\textbf{解}\quad
$$\begin{aligned} y^{\prime}=e^{-x^{2}}(-2 x) ,y^{\prime \prime} &=e^{-x^{2}}(-2 x)^{2}+(-2) e^{-x^{2}}=e^{-x^{2}}\left(4 x^{2}-2\right) \end{aligned}$$

\subsection{$y=\left(1+x^{2}\right)$}
\textbf{解}\quad
$$y^{\prime}=2 x, y^{\prime\prime}=2$$

\textbf{\textcolor{red}{注}}\quad
求一次导数以后幂次会-1,可以通过这个性质来粗略估算自己有没有算错,例如第1.1是$x$的二次的函数,则求两次导数以后应该是$0$次的.$\ln x$也可以算是$0$次的.其阶数小于$x^{\alpha},0<\alpha<1$

\section{求下列函数的 $n$ 阶导数:}
\subsection{$y=\frac{x^{2}}{x^{2}-1}$}
\textbf{解}\quad
$y=1+\frac{1}{x^{2}-1}=1+\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)$

于是易知$$y^{(n)}=\frac{n !}{2}\left((x-1)^{-(n+1)}-(x+1)^{-(n+1)}\right) \cdot(-1)^{n}$$

\subsection{$y=\frac{1}{x^{2}-3 x+2}$}
\textbf{解}\quad
$y=\frac{1}{(x-2)(x-1)}=\frac{1}{x-2}-\frac{1}{x-1}$

于是易知$$y^{(n)}=n !\left((x-2)^{-(n+1)}-(x-1)^{-(n+1)}\right)(-1)^{n}$$

\subsection{$y=\sin ^{4} x+\cos ^{4} x$}
\textbf{解}\quad
$$y=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x=1-\frac{1}{2} \sin ^{2} 2 x=1-\frac{1}{4}(1-\cos 4x)=\frac{3}{4}+\frac{1}{4}\cos 4x$$

于是易知$$y^{(n)}=4^{n-1} \cos \left(4 x+\frac{n \pi}{2}\right)$$

\subsection{$y=\frac{a x+b}{c x+d}$}
\textbf{解}\quad
(i) $c \neq 0, y=\frac{a}{c}+\frac{b-\frac{a}{c} d}{c x+d}=\frac{a}{c}+\frac{\frac{b}{c}-\frac{a}{c^{2}} d}{x+\frac{d}{c}}=\frac{a}{c}+\frac{l}{x+\frac{d}{c}}, l=\frac{b}{c}-\frac{a}{c^{2}} d$

$y^{(n)}=l \cdot \sum_{k=0}^{n} \frac{(-1)^{k} k !}{\left(x+\frac{d}{c}\right)^{k+1}}=\frac{c b-a d}{c^{2}} \cdot \sum_{k=0}^{n} \frac{(-1)^{k} k !}{\left(x+\frac{d}{c}\right)^{k+1}}$

(ii) $c=0, d \neq 0, y=\frac{a}{d} x+\frac{b}{d}, y^{\prime}=\frac{a}{d}, y^{(n)}=0, n=2,3, \cdots$

\subsection{$y=\sin a x \sin b x$}
\textbf{解}\quad
$\begin{aligned} y &=\sin a x \sin b x=-\frac{1}{2}[\cos (a+b) x-\cos (a-b) x] \\ \text{由此可知},y^{(n)} &=\frac{1}{2}[\cos (a-b) x]^{(n)}-\frac{1}{2}[\cos (a+b) x]^{(n)} \\ &=\frac{1}{2}(a-b)^{n} \cos \left[(a-b) x+\frac{n \pi}{2}\right]-\frac{1}{2}(a+b)^{n} \cos \left[(a+b) x+\frac{n \pi}{2}\right] \end{aligned}$

\subsection{$y=\frac{1+x}{\sqrt{1-x}}$}
\textbf{解}\quad
由莱布尼茨公式:$y^{(n)}=(1+x)\left(\frac{1}{\sqrt{1-x}}\right)^{(n)}+n\left(\frac{1}{\sqrt{1-x}}\right)^{(n-1)}$

$\begin{aligned}\left(\frac{1}{\sqrt{1-x}}\right)^{(n)} &=\left[(1-x)^{-\frac{1}{2}}\right]^{(n)} \\ &=(-1)^{n} \cdot-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right) \cdots\left(-\frac{1}{2}-n+1\right)(1-x)^{-\frac{1}{2}-n} \\ &=\frac{(2 n-1) ! !}{2^{n}}(1-x)^{-\frac{1}{2}-n} \\\left[(1-x)^{-\frac{1}{2}}\right]^{n-1} &=\frac{(2 n-3) ! !}{2^{n-1}}(1-x)^{-\frac{1}{2}-n+1} \end{aligned}$

$\begin{aligned} y^{(n)} &=\frac{(2 n-1) ! !}{2^{n}}(1-x)^{-\frac{1}{2}-n}(1+x)+n \frac{(2 n-3) ! !}{2^{n-1}}(1-x)^{-\frac{1}{2}-n+1} \\ &=\frac{(2 n-3) ! !}{2^{n}}(1-x)^{-\frac{1}{2}-n}(4 n-x-1) \end{aligned}$

\subsection{$y=\mathrm{e}^{x}(\sin x+\cos x)$}
\textbf{解}\quad
$y=\mathrm{e}^{x}(\sin x+\cos x)=\sqrt{2} \mathrm{e}^{x} \sin \left(x+\frac{\pi}{4}\right)$

归纳:设$n=k$时满足$y^{(k)}=(\sqrt{2})^{k+1} \mathrm{e}^{x} \sin \left(x+\frac{(k+1) \pi}{4}\right)$

则$n=k+1$时,有
$$
\begin{aligned} y^{(k+1)} &=\sqrt{2}^{k+1} e^{x} \sin \left(x+\frac{(n+1) \pi}{4}\right)+\sqrt{2}^{k+1} e^{x} \cos \left(x+\frac{(n+1) \pi}{4}\right) \\
	&=\sqrt{2}^{k+2} e^{x} \sin \left(x+\frac{(n+2) \pi}{4}\right) \end{aligned}
$$

归纳成立.

\subsection{$y=\ln \frac{a+b x}{a-b x}$}
\textbf{解}\quad
$$y=\ln (a+b x)-\ln (a-b x)$$

$$[\ln (a+b x)]^{(n)}=(-1)^{n-1} b^{n}(n-1) !(a+b x)^{-n}$$

$$[\ln (a-b x)]^{(n)} =(-1)^{n-1}(-b)^{n}(n-1) !(a-b x)^{-n} =-b^{n}(n-1) !(a-b x)^{-n} $$

$$y^{(n)}=(n-1) ! b^{n}\left[\frac{(-1)^{n-1}}{(a+b x)^{n}}-\frac{1}{(a-b x)^{n}}\right]$$

\section{求下列函数的 $n$ 阶导数 :}
\subsection{$y=x^{2} \mathrm{e}^{3 x}$}
\textbf{解}\quad
由莱布尼茨公式:

$$
	\begin{aligned} y^{(n)} & =x^{2}\left(e^{3 x}\right)^{(n)}+C_{n}^{1} 2 x\left(e^{3 x}\right)^{(n-1)}+\ln ^{2} 2\left(e^{3 x}\right)^{(n-2)} \\
		        & =3^{n} x^{2} e^{3 x}+3^{n-1} 2 n x e^{3 x}+n(n-1) 3^{n-2} e^{3 x}
	\end{aligned}
$$

\subsection{$y=x^{2} \sin 3 x$}
\textbf{解}\quad
由莱布尼茨公式:

$$
	\begin{aligned}
		y^{(n)} & =x^{2}(\sin 3 x)^{(n)}+2 n x(\sin 3 x)^{(n-1)}+n(n-1)(\sin 3 x)^{(n-2)}                                                                                             \\
		        & =3^{n} x^{2} \sin \left(3 x+\frac{n \pi}{2}\right)+2 n x 3^{n-1} \sin \left[3 x+\frac{(n-1) \pi}{2}\right]+n(n-1) 3^{n-2} \sin \left[3 x+\frac{(n-2) \pi}{2}\right]
	\end{aligned}
$$

\subsection{$y=\left(2 x^{2}+1\right) \sinh x$}
\textbf{解}\quad
由莱布尼茨公式:

$$
	\begin{aligned} y^{(n)} & = \left(2 x^{2}+1\right)(\sinh x)^{(n)}+4 n x(\sinh x)^{(n-1)}+2 n(n-1)(\sinh x)^{(n-2)}                                                                                                                                    \\
		        & =\left(2 x^{2}+1\right)\left(\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}\right)^{(n)}+4 n x\left(\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}\right)^{(n-1)}+2 n(n-1)\left(\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}\right)^{(n-2)} \\
		        & = \frac{\left(2 x^{2}+1\right)}{2}\left[\mathrm{e}^{x}-(-1)^{n} \mathrm{e}^{-x}\right]+2 n x\left[\mathrm{e}^{x}-(-1)^{n-1} \mathrm{e}^{-x}\right]+ n(n-1)\left[\mathrm{e}^{x}-(-1)^{n-2} \mathrm{e}^{-x}\right]
	\end{aligned}
$$

\section{对下列方程所确定的隐函数 $y=y(x),$ 求 $\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}$ :}
\subsection{$\tan (x+y)-x y=0$}
\textbf{解}\quad
$$\sec ^{2}(x+y)\left(1+y^{\prime}\right)-y-x y^{\prime}=0$$

解得
$$y^{\prime}=\frac{y-\sec ^{2}(x+y)}{\sec ^{2}(x+y)-x}$$

对上式求导,得

$$
	\begin{aligned}y^{\prime \prime} & =\frac{2 y^{\prime}-2 \sec ^{2}(x+y) \tan (x+y)\left(1+y^{\prime}\right)^{2}}{\sec ^{2}(x+y)-x}                                                               \\
		                  & =\frac{2 \frac{y-\sec ^{2}(x+y)}{\sec ^{2}(x+y)-x}-2 \sec ^{2}(x+y) \tan (x+y)\left[1+\frac{y-\sec ^{2}(x+y)}{\sec ^{2}(x+y)-x}\right]^{2}}{\sec ^{2}(x+y)-x}\end{aligned}
$$

\subsection{$x^{3}+y^{3}-3 a x y=0$}
\textbf{解}\quad
对左侧关于$x$求导,得

$$\mathrm{LHS'}=3 x^{2}+3 y^{2} y^{\prime}-3 a y-3 a x y^{\prime}=0, \quad y^{\prime}=\frac{x^{2}-a y}{a x-y^{2}}$$

对上式求导,得

$$y^{\prime \prime}=\frac{2 x+2 y y^{\prime}-2 a y^{\prime}}{a x-y^{2}}=\frac{2 x+2(y-a) \frac{x^{2}-a y}{a x-y^{2}}}{a x-y^{2}}$$

\section{对下列参数方程确定的函数 $y=y(x),$ 求 $\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}$ :}
\subsection{$\left\{\begin{array}{l}x=a t^{2} \\ y=a t^{3}\end{array}\right.$}
\textbf{解}\quad
$$y^{\prime}=\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{3 a t^2}{2 a t}=\frac{3at}{2}$$
$$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=\frac{\mathrm{d} y^{\prime}(x)}{\mathrm{d} x}=\frac{3}{4 a t}$$

\subsection{$\left\{\begin{array}{l}x=a t \cos t \\ y=a t \sin t\end{array}\right.$}
\textbf{解}\quad
$$y^{\prime}=\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{a \sin t+a t \cos t}{a \cos t-a t \sin t}=\frac{\sin t+t \cos t}{\cos t-t \sin t}$$
$$\begin{aligned} \frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} &=\frac{\mathrm{d} y^{\prime}(x)}{\mathrm{d} x}\\&=\frac{\left(\frac{\sin t+t \cos t}{\cos t-t \sin t}\right)^\prime}{a \cos t-a t \sin t} \\ &=\frac{(2 \cos t-t \sin t)(a \cos t-a t \sin t)-(\sin t+t \cos t) a(-2 \sin t-t \cos t)}{(a \cos t-a t \sin t)^{3}} \\ &=\frac{2+t^{2}}{a(\cos t-t \sin t)^{3}} \end{aligned}$$

\section{设 $y=(\arcsin x)^{2}$}
\subsection{证明 :$\left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}=2$}
\begin{proof}
	$$y^{\prime}=2(\arcsin x)(\arcsin x)^{\prime}=2(\arcsin x) \frac{1}{\sqrt{1-x^{2}}}$$
	$$y^{\prime \prime}=2(\arcsin x)^{\prime} \frac{1}{\sqrt{1-x^{2}}}+2(\arcsin x)\left(\frac{1}{\sqrt{1-x^{2}}}\right)^{\prime}=\frac{2}{1-x^{2}}+2(\arcsin x) \frac{x}{{\left(1-x^{2}\right)^\frac{3}{2}}}$$
	$$\left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}=2+2 \frac{x}{\sqrt{1-x^{2}}}(\arcsin x)-2 \frac{x}{\sqrt{1-x^{2}}}(\arcsin x)=2$$
\end{proof}

\subsection{求 $y^{(n)}(0)$}
\textbf{解}\quad
$y^{\prime}(0)=2(\arcsin 0) \frac{1}{\sqrt{1-0^{2}}}=0$,$y^{\prime \prime}(0)=2$

对等式 $\left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}=2$ 两端求 $n-2$ 阶导数,并将$x=0$带入得：

$y^{(n)}-(n-2)^{2} y^{(n-2)}=0$

得$y^{(2 k)}(0)=(2 k)^{2} y^{(2 k-2)}(0)=(2 k-2)^{2}(2 k-4)^{2} y^{(2 k-4)}(0)=\cdots=(2 k-2) ! ! y^{\prime \prime}(0)=2^{k}(k-1) !$

$$y^{(2 k+1)}(0)=(2 k-1)^{2} y^{(2 k-1)}(0)=(2 k-1)^{2}(2 k-3)^{2} y^{(2 k-3)}(0)=\cdots=(2 k-1) ! ! y^{\prime}(0)=0$$

\section{证明: 切比雪夫多项式 $T_{n}(x)=\frac{1}{2^{n-1}} \cos (n \arccos x)$ 满足方程
$$
	\left(1-x^{2}\right) T_{n}^{\prime \prime}(x)-x T_{n}^{\prime}(x)+n^{2} T_{n}(x)=0
$$}
\begin{proof}
	$$\begin{aligned} 2^{n-1} T_{n}^{\prime}(x) &=[\cos (n \arccos x)]^{\prime}=-\sin (n \arccos x) \cdot n(\arccos x)^{\prime} \\ &=n \sin (n \arccos x) \cdot \frac{1}{\sqrt{1-x^{2}}} \end{aligned}$$
	$$
		\begin{aligned}
			2^{n-1} T_{n}^{\prime \prime}(x) & =\left[n \sin (n \arccos x) \cdot \frac{1}{\sqrt{1-x^{2}}}\right]^{\prime}                                                                        \\
											 & =n\left[\cos (n \arccos x) \cdot\left(\frac{1}{\sqrt{1-x^{2}}}\right)^{2}+\sin (n \arccos x)\left(\frac{1}{\sqrt{1-x^{2}}}\right)^{\prime}\right] \\
											 & =n\left[\cos (n \arccos x) \cdot \frac{n}{1-x^{2}}+\sin (n \arccos x) \frac{x}{\sqrt{\left(1-x^{2}\right)^{3}}}\right]                            \\
			2^{n-1} T_{n}(x)=                & \cos (n \arccos x)
		\end{aligned}
	$$
	
	三个式子加和,可得 $\left(1-x^{2}\right) T_{n}^{\prime \prime}(x)-x T_{n}^{\prime}(x)+n^{2} T_{n}(x)=0 .$		
\end{proof}

\section{证明函数$f(x)=\left\{\begin{array}{ll}
		  x^{2 n} \sin \frac{1}{x}, & x \neq 0 \\
		  0,                        & x=0\end{array}\right.$在零点有 $n$ 阶导数,但无 $n+1$ 阶导数 ($n$ 为正整数).}
\begin{proof}
	由莱布尼茨公式,当 $x \neq 0$ 时得
	$$
		f^{(m)}(x)=\left(x^{2 n} \sin \frac{1}{x}\right)^{(m)}=\sum_{i=0}^{m} C_{m}^{i}\left(x^{2 n}\right)^{(m-i)}\left(\sin \frac{1}{x}\right)^{(i)}
	$$
	
	首先指出,有
	$$
		\left(\sin \frac{1}{x}\right)^{(i)}=\sum_{k=1}^{i-1}\left[a_{k} x^{-(i+k)} \sin \left(\frac{1}{x}+\frac{k \pi}{2}\right)\right]+\left(-x^{-2}\right)^{i} \sin \left(\frac{1}{x}+\frac{i \pi}{2}\right) \quad(x \neq 0)
	$$
	
	其中 $a_{k}$ 是某些常数. 现用数学归纳法予以证明:
	
	当 $i=1$ 时,命题显然成立;
	
	设当 $i=N$ 时,命题成立,要证命题对 $i=N+1$ 时也成立.事实上,有
	$$
		\begin{aligned}
			  & \left(\sin \frac{1}{x}\right)^{(N+1)}=\sum_{k=1}^{N-1} a_{k}\left[x^{-(N+k)} \sin \left(\frac{1}{x}+\frac{k \pi}{2}\right)\right]^{\prime}+\left[\left(-x^{-2}\right)^{N} \sin \left(\frac{1}{x}+\frac{N \pi}{2}\right)\right]^{\prime} \\
			= & \sum_{k=1}^{N-1} a_{k}\left[-(N+k) x^{-(N+1+k)} \sin \left(\frac{1}{x}+\frac{k \pi}{2}\right)+x^{-(N+k)}\left(-x^{-2}\right) \sin \left(\frac{1}{x}+\frac{k+1}{2} \pi\right)\right]                                                     \\
			  & +\left[N\left(-x^{-2}\right)^{N-1}\left(2 x^{-3}\right) \sin \left(\frac{1}{x}+\frac{N \pi}{2}\right)+\left(-x^{-2}\right)^{N+1} \sin \left(\frac{1}{x}+\frac{N+1}{2} \pi\right)\right]                                                 \\
			= & \sum_{k=1}^{(N+1)-1}\left[b_{k} x^{-(N+1+k)} \sin \left(\frac{1}{x}+\frac{k \pi}{2}\right)\right]+\left(-x^{-2}\right)^{N+1} \sin \left(\frac{1}{x}+\frac{N+1}{2} \pi\right)
		\end{aligned}
	$$
	
	因而,我们有
	$$
		f^{(m)}(x)=\sum_{i=0}^{m} C_{m}^{i} \frac{(2 n) !}{(2 n-m+i) !} x^{2 n-m+i}\left[\sum_{k=1}^{i-1} a_{k} x^{-(i+k)} \sin \left(\frac{1}{x}+\frac{k \pi}{2}\right)+\left(-x^{-2}\right)^{i} \sin \left(\frac{1}{x}+\frac{i \pi}{2}\right)\right]
	$$
	
	于是,
	$$
		f^{(m)}(x)=(-1)^{m} x^{2(n-m)} \sin \left(\frac{1}{x}+\frac{m \pi}{2}\right)+O\left(|x|^{2(n-m)+1}\right) \quad(x \rightarrow 0) \quad(m=1,2, \cdots, n)
	$$
	
	由于
	$$
		f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}=\lim _{x \rightarrow 0} \frac{f(x)}{x}=\lim _{x \rightarrow 0} x^{2 n-1} \sin \frac{1}{x}=0
	$$
	
	故由（*）式,得知
	$$
		f^{\prime \prime}(0)=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}=\lim _{x \rightarrow 0}\left[-x^{2 n-3} \sin \left(\frac{1}{x}+\frac{\pi}{2}\right)+O\left(|x|^{2 n-2}\right)\right]=0
	$$
	
	一直推下去,得
	$$
		f^{(n)}(0)=\lim _{x \rightarrow 0} \frac{f^{(n-1)}(x)}{x}=\lim _{x \rightarrow 0}\left[(-1)^{n-1} x \sin \left(\frac{1}{x}+\frac{(n-1) \pi}{2}\right)+O\left(|x|^{2}\right)\right]=0
	$$
	
	但
	$$
		f^{(n+1)}(0)=\lim _{x \rightarrow 0} \frac{f^{(n)}(x)}{x}=\lim _{x \rightarrow 0}\left[\frac{(-1)^{n}}{x} \sin \left(\frac{1}{x}+\frac{n \pi}{2}\right)+O(1)\right]
	$$
	
	在 $x=0$ 附近 $, \frac{(-1)^{n}}{x} \sin \left(\frac{1}{x}+\frac{n \pi}{2}\right)$ 无界且振荡,故
	$$
		\lim _{x \rightarrow 0}\left[\frac{(-1)^{n}}{x} \sin \left(\frac{1}{x}+\frac{n \pi}{2}\right)+O(1)\right]
	$$
	
	不存在,因而 $f^{(n+1)}$ (0)不存在.
\end{proof}
\begin{remark}
	本题的主要训练目的在于充分了解在某处导数和在整体区间上的导函数之间的区别和联系.一般来说,如果在整个区间上每个点处都存在导数,那么导数值在区间上构成的函数被称为导函数.在本题中, $f(x)$在$(0,\infty)$上存在导函数,而且可以求任意阶导数.但是我们不可以先求出导函数$f'(x)$,然后令$x\rightarrow 0$,或者直接取$x=0$,得到在$x=0$处的导数,因为导函数在该点处并不连续(第二类间断点).所以整体思路应是这样:

	研究$f(x)$在$(0,\infty)$上的$m$阶导函数的表达式的形式,然后利用归纳法,(先猜出在$x=0$处的前$2n-1$阶导数都为$0$)证明小于$2n$阶时导数都为$0$.这一步需要使用定义:$f^{(m+1)}(x)=\dfrac{f^{(m)(x)-f^{(m)}(0)}}{x-0}$.
\end{remark}


\section{设函数 $f(x)=\left\{\begin{array}{ll}x^{2 n} \sin \frac{1}{x}, & x \neq 0 ,\text { ,求 } f^{(n)} \\ 0, & x=0\end{array}\right.$}
\textbf{解}\quad
根据上题结论可知 $f^{(n)}(0)=0 .$

\section{设函数 $f(x)$ 是可以求任意阶导数的函数,使用数学归纳法证明:
  $$
	  \left[x^{n-1} f\left(\frac{1}{x}\right)\right]^{(n)}=\frac{(-1)^{n}}{x^{n+1}} f^{(n)}\left(\frac{1}{x}\right), \quad n=1,2, \cdots
  $$}
\begin{proof}
	$n=1$时显然成立.

	假设 $n \leqslant k$ 时结论也成立,则当 $n=k+1$ 时,
	
	$$
		\begin{aligned}
			\left[x^{k} f\left(\frac{1}{x}\right)\right]^{(k+1)} & =\left[k x^{k-1} f\left(\frac{1}{x}\right)+x^{k} f^{\prime}\left(\frac{1}{x}\right)\left(\frac{1}{x}\right)^{\prime}\right]^{(k)}                                                                                                       \\
																 & =k\left[x^{k-1} f\left(\frac{1}{x}\right)\right]^{k}-\left[x^{k-2} f^{\prime}\left({\frac{1}{x}}\right)\right]^{k}                                                                                                                      \\
																 & =k\left[\frac{(-1)^{k}}{x^{k+1}} f^{(k)}\left(\frac{1}{x}\right)\right]-\left[\frac{(-1)^{k-1}}{x^{k}} f^{(k)}\left(\frac{1}{x}\right)\right]^{\prime}                                                                                  \\
																 & =k\left[\frac{(-1)^{k}}{x^{k+1}} f^{(k)}\left(\frac{1}{x}\right)\right]-\left[k \frac{(-1)^{k}}{x^{k+1}} f^{(k)}\left(\frac{1}{x}\right)+\frac{(-1)^{k-1}}{x^{k}} f^{(k+1)}\left(\frac{1}{x}\right)\left(-\frac{1}{x^{2}}\right)\right] \\
																 & =\frac{(-1)^{k+1}}{x^{k+2}} f^{(k+1)}\left(\frac{1}{x}\right)
		\end{aligned}
	$$
	归纳成立.		
\end{proof}

\end{document}
